Equilateral triangle ABC is inscribed in a circle. The angle bisector at A intersects the circle again at P (AP is the diameter). If the length of segment AC is 10, find AP - (BP + CP).
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Another approach . . .
Equilateral triangle is inscribed in a circle.
The angle bisector at intersects the circle again at ( is adiameter).
If the length of segment is 10, find: .Code:A * o * * /|\ * * / | \ * * / | \ 10 * / |30°\ * / | \ * * / | \ * * / | \ * / | \ B o- - - - + - - - -o C * | * * * 60° | * * * o * P
We have: inscribed equilateral triangle
. . with diameter
Draw chord .
is inscribed in a semicircle.
. . Hence: .
We have: . .(30°-60° right triangle)
Pythagorus says: .